Physics Procedia 00 (2009) 1–9

Physics Procedia

A Vertex-Number-Evolving Markov Chain of Networks1

Dinghua Shia, Hong Xua and Liming Liub2

aDepartment of Mathematics, Shanghai University, Shanghai 200444, ChinabDepartment of Logistics and Maritime Studies, Hong Kong Polytechnic University, Hong Kong, China

Abstract

We have introduced a vector Markov chain of the vertex number with degree k in networkevolving process as a framework of theoretical analysis and proved the stability of the BA-1model and the LCD-1 model. In this paper, we use the vertex-number-evolving Markov chainto prove rigorously the existence of the steady-state degree distribution P(k) for a special caseof the initial attraction model allowing multiple edges. The application of our approach to theLCD-m model, the result shows that it is more simpler than Bollobas’ method.

Keywords: scale-free network, BA model, LCD model, initial attraction model, vertex numberwith degree k, Markov chain, stability, convergence in probabilityPACS: 89.75.Hc, 02.50.Ga, 89.75.Da

1. Introduction

Barabasi and Albert [1] observed that for many real networks, the fraction P(k) of verticeswith degree k is proportional over a large range to a “scale-free” power-law tail: k−γ, where γ isa constant independent of the size of the network. To explain this phenomenon, they proposedthe following network-generating mechanism, known as the BA model:

“· · · starting with a small number (m0) of vertices, at every time step we add a new vertexwith m (≤ m0) edges that link the new vertex to m different vertices already present in the system.To incorporate preferential attachment, we assume that the probability Π that a new vertex willbe connected to a vertex depends on the connectivity ki of that vertex, so that Π(ki) = ki/

∑j k j.

After t steps the model leads to a random network with t + m0 vertices and mt edges.”This discovery of the scale-free topology in complex networks has been followed by intensive

research in constructing models to predict and analyze network topological structures in growing

1Supported by the National Natural Science Foundation of China through grants 60874083 and 10872119, and byHong Kong RGC through grant PolyU5226/07E.

2Corresponding author. Email: shidh2001@263.net or lgtliulm@polyu.edu.hk

c© 2010 Published by Elsevier Ltd

Physics Procedia 3 (2010) 1757–1765

www.elsevier.com/locate/procedia

1875-3892 c© 2010 Published by Elsevier Ltddoi:10.1016/j.phpro.2010.07.016

D. H. Shi et al. / Physics Procedia 00 (2009) 1–9 2

networks, e.g., [2, 3, 4, 5].However, for the network degree distribution, statistics of real networks and simulations of

many modelled networks suffer from finite size effect. In the literature of complex networksthere are some examples [6, 7] showing that simulation results of modelled networks seem tofollow power laws at any given time step, but their limits are actually not. Various methodsto predict the power law of modelled networks depends on the thermodynamic limit. Doesthe thermodynamic limit for a modelled networks exist? To answer this question convincingly,rigorous mathematical proof is needed. In this paper, we present an effective approach to provethe existence of thermodynamic limit for an initial attraction model.

We organize the paper as follows. In Section 2, we first review some of the existing methodsfor network degree distributions. In Section 3, we present a new approach to prove the existenceof the steady-state degree distribution for a special case of the initial attraction model allowingmultiple edges. In Section 4, we show how to use the new approach to the LCD model andconclude the paper and future research opportunities.

2. Existing methods

Since the BA model was proposed, there have been three methods to derive its degree distri-bution in literature. These methods either approximated or assumed the existence of P(k), hencethey are only heuristics. Bollobas et al. [4] first proved the existence of P(k) for the LCD model.Recently, Hou et al. [8] proved the existence of P(k) for the HK model with p = 1. Here, wefirst briefly describe these methods before presenting our new method.

Barabasi et al. [9] considered the dynamic equation of the degree ki(t) of vertex i chosenrandomly at time t to analyze the BA model. Using a mean-field like approximation argument,i.e., assuming that the probability of an existing vertex i received an new edge is mΠ(ki) andvertex i is uniformly distributed in t vertices, they obtained the solution

P(k) ∼ 2m2k−3.

Here, γ = 3 is very closely to the simulation result γ � 2.9 ± 0.1.Krapivsky et al. [2] replaced the degree ki(t) of vertex i at time t by the vertex number Nk(t)

with degree k in the whole network at time t, thereby established its rate equation. Assuming thatthe steady-state degree distribution P(k) exists and Nk(t)/t converges to P(k) in probability, theygave the exact solution of P(k) for the BA model with m = 1 as follows:

P(k) =4

k(k + 1)(k + 2).

Dorogovtsev et al. [3] regarded the degree ki(t) at time t of the vertex added at time step i asa random variable, and investigated its distribution P(k, i, t) = P{ki(t) = k}, thereby established aset of master equations. Ignoring the vertices in the initial network, let P(k, t) = 1

t

∑ti=1 P(k, i, t).

Assuming that the steady-state degree distribution P(k) exists and limt→∞ t[P(k, t+1)−P(k, t)] =0, they gave the exact solution of P(k) for the BA model allowing multiple edges as follows:

P(k) =2m(m + 1)

k(k + 1)(k + 2).

However, Bollobas and Riordan [10] made a general comment on the BA model and theabove related works. They point out: “from a mathematical point of view, the description above

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D. H. Shi et al. / Physics Procedia 00 (2009) 1–9 3

does not make sense. The first problem is getting started. The second problem is with thepreferential attachment rule itself, and arises only for m ≥ 2. In order to prove results about theBA model, one must first decide on the details of the model itself.”

To make the BA model more operable, Bollobas et al. [4] recommended a Linearized ChordDiagram (LCD-m) model. The LCD-m model allows multiple edges and loops, so it is notidentical to the BA model even when m = 1.

The LCD-1 model starts with the graph G1 with one vertex and one loop. Given Gt, the graphGt+1 is obtained form Gt by adding the vertex t together with a single edge directed from t to i,where i is chosen randomly with preferential probability

Π(i = s) =

{ks/(2t + 1), 1 ≤ s ≤ t − 11/(2t + 1), s = t

, (1)

where ks is the degree of the vertex s at time t.For the LCD-m model with m > 1, we define the process {Gt

m}t≥0 by running the process {Gt1}

on a sequence v′1, v′2, · · ·; the graph Gtm is formed from by identifying the vertices v′1, v′2, · · · , v′m

to form v1, identifying v′m+1, v′m+2, · · · , v′2m to form v2, and so on.Let Nk(t) be the number of vertices with (total) degree k = m+q (where q is in-degree) at time

t for the model, based on the martingale theory, they rigorously proved that Nk(t)/t converges toP(k) in probability.

In order to incorporate high clustering, Holme and Kim [5] proposed a tunable clustering BAmodel. Consider the case of p = 1 in the model and called HK-p = 1 model: the first edgeconnects to an existing vertex in the same way as the BA model; the remaining m − 1 edges arerandomly chosen from inside the neighborhood of the chosen vertex without allowing multipleedges. Thus the probability of an existing vertex i received an new edge is

Πm(ki) =ki∑j k j+∑

l∈Oi

kl∑j k j

m − 1kl= mΠ(ki). (2)

Although it is now exact, but when m ≥ 2, the way how the edges are connected has changed.Recently, using the first-passage probability in the Markov chain theory, Hou et al. [8]

rigorously proved the existence of the steady-state degree distribution for the HK-p = 1 model.The proof in [8] used a set of non-homogenous Markov chains based on the vertex degree ki(t),which was first proposed by Shi et al. [11].

3. New approach

Xu and Shi [12] introduced a vector Markov chain of the vertex number with degree k innetwork evolving process, brief called the vertex-number-evolving Markov chain. Now we usethe Markov chain rigorously to prove the existence of the steady-state degree distribution P(k)for a special case of the initial attraction model allowing multiple edges.

Consider the following D-w.m.e model. When t = 1, the initial network consists of twovertices and m multiple edges. Then at every time step, we add a new vertex with m edgesconnecting to the existing network. We assume that a vertex i will receive l edges from a newvertex according to the degree-preferential rule, i.e., the probability Πl

m(ki) of chosen vertex idepends on its degree ki, so that

Πlm(ki) = Cl

m

⎛⎜⎜⎜⎜⎜⎜⎝ki/∑

j

k j

⎞⎟⎟⎟⎟⎟⎟⎠l ⎛⎜⎜⎜⎜⎜⎜⎝1 − ki/

∑

j

k j

⎞⎟⎟⎟⎟⎟⎟⎠m−l

. (3)

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D. H. Shi et al. / Physics Procedia 00 (2009) 1–9 4

The D-w.m.e model is a special case of the initial attraction model allowing multiple edgesproposed by Dorogovtsev et al. [3].

The D-w.m.e model shows that if the network Gt at time step t is given, the probabilisticlaw of the network Gt+1 at time step t + 1 can be completely determined by network-generatingmechanism. Now, we define a vector Markov chain N(t) = {Nm(t), · · · ,Nk(t), · · ·} where Nk(t) isthe vertex number with degree k in the network Gt. Obviously, Nk(t) ≡ 0 for all k > mt.

Using the mathematical expectation E[N(t)] of N(t) (component-wise), we may define thetransient degree distribution of the network Gt as P(t) = E[N(t)]/(t+1) = {P(m, t), · · · , P(k, t), · · ·}.If limt→∞ P(t) = {P(m), · · · , P(k), · · ·} = P exists in component-wise, the degree sequence P iscalled the steady-state degree distribution of the network.

First, we consider the incremental process Y(t) = N(t + 1) − N(t). When a vertex in Nk(t)received an edge at time step t, it implies that Nk(t + 1) minus 1, but Nk+1(t + 1) will increaseby 1. In addition, as a new vertex is added, Nm(t + 1) will increase by 1. Hence it is clear thatYm(t) − 1 can only take the values of {0,−1, · · · ,−m} and every other Yk(t) all can take the valuesof {−m, · · · ,−1, 0, 1, · · · ,m} for the BAw.m.e model. According to the degree-preferential rule,we have

Lemma 1. For the D-w.m.e model, let pk =k

2mt and αk = pkNk(t), the condition distributionsof Ym(t) and Yk(t) can be represented by αk as follows respectively.

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

P{Ym(t) − 1 = −l|N(t)} = Clmα

lm(1 − αm)m−l + O(1/t)

P{Yk(t) = l|N(t)} = ∑ml

j=0 Cl+ jm C j

m−l− jαl+ jk α

jk−1(1 − αk − αk−1)m−l−2 j + O(1/t)

P{Yk(t) = −l|N(t)} = ∑ml

j=0 Cl+ jm C j

m−l− jαl+ jk α

jk−1(1 − αk − αk−1)m−l−2 j + O(1/t)

, (4)

where l = 0,−1, · · · ,−m and ml = [(m − l)/2].Proof. The reason why formula (4) holds is as follows. Once the network Gt is given, the

total degrees in the network are 2mt. When an edge is added to the network, the probability thatthe vertex with degree k received it is obviously pk =

k2mt = O(1/t) according to the degree-

preferential rule. Thus, the probability of Nk(t) increased by 1 is just αk = pkNk(t) < 1. Inaddition, when m > 1 edges are added to the network, by formula (3), the probability that avertex received more than an edge is O(1/t). In the following arguments, we may ignore casesof a vertex connecting multiple edges provided that there is the term O(1/t) in the result.

Now for the event {Ym(t) − 1 = −l|N(t)}, we only needs to consider the case when there arel edges connecting to l different vertices with degree m in new added m edges and other m − ledges all are not connecting to vertices with degree m. Hence, P{Ym(t) − 1 = l|N(t)} should beCl

mXm(t)pm · · · [Xm(t) − l + 1] pm(1 − αm)m−l = Clmα

lm(1 − αm)m−l + O(1/t).

When m+ 1 ≤ k ≤ 2mt, for the event {Yk(t) = l|N(t)}, we only needs to discuss the case whenthere are l+ j edges connecting to l+ j different vertices with degree k−1 and j edges connectingto j different vertices with degree k in new added m edges, and other m − l − 2 j edges all are notconnecting to vertices with degrees k − 1 and k. Obviously, l + 2 j ≤ m must hold, it implies that0 ≤ j ≤ [(m − l)/2] = ml. Thus we obtain the second equality in (4). Using symmetry, we mayobtain the third equality in (4).

Next, based on formulas (4) we have the following key lemma.Lemma 2. For the D-w.m.e model, the mathematical expectation of Ym(t) and Yk(t) may be

represented as follows, respectively.{

E[Ym(t)] = 1 − mpmE[Nm(t)] + O(1/t) = 1 − mαm + O(1/t)E[Yk(t)] = m (pk−1E[Nk−1(t)] − pkE[Nk(t)]) + O(1/t)

. (5)

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D. H. Shi et al. / Physics Procedia 00 (2009) 1–9 5

Proof. Using the definition of conditional expectation, taking the expectation for the randomvariable {Ym(t) − 1 = −l|N(t)} and noting

∑ml=1 lCl

mαlm(1 − αm)m−l = mαm and E[Ym(t) − 1|N(t)] =

E[Ym(t)|N(t)] − 1, we have from the formula (4)

E[Ym(t)|N(t)] = 1 −m∑

l=1

lClmα

lm(1 − αm)m−l + O(1/t) = 1 − mαm + O(1/t). (6)

Again, by the property of conditional expectation, i.e., E[E[Ym(t)−1|N(t)]] = E[Ym(t)−1], takingthe expectation on both sides of the above expression we obtain the first equality in (5).

Similarly, for Yk(t) we have from formula (4)E[Yk(t)|N(t)] =

∑ml=1 l [P{Yk(t) = l|N(t)} − P{Yk(t) = −l|X(t)}]

=∑m

l=1 l∑ml

j=0 Cl+ jm C j

m−l− j(αl+ jk−1α

jk − αl+ j

k αjk−1)(1 − αk − αk−1)m−l−2 j + O(1/t)

=∑m

l=1 l∑ml

j=1 Cl+ jm C j

m−l− j(αl+ jk−1α

jk − αl+ j

k αjk−1)(1 − αk − αk−1)m−l−2 j

+∑m

l=1 lClm(αl

k−1 − αlk) × (1 − αk − αk−1)m−l + O(1/t)

=∑m

l=1 l∑ml

j=1 Cl+ jm C j

m−l− j(αl+ jk−1α

jk − αl+ j

k αjk−1)(1 − αk − αk−1)m−l−2 j

+ m[αk−1(1 − αk)m−1 − αk(1 − αk−1)m−1] + O(1/t)

= m(αk−1 − αk) + O(1/t). (7)

The proof of the last equality in (7) is given in the Appendix. Again taking the expectationon both sides of the above expression we obtain the second equality in (5). Thus the proof ofLemma 2 is completed.

Formula (5) can be rewritten as the following system of difference equations:⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

E[Nm(t + 1)] − E[Nm(t)] + m2t E[Nm(t)] = 1 + O(1/t),

E[Nj(t + 1)] − E[Nj(t)] +j

2t E[Nj(t)]=

( j−1)2t E[Nj−1(t)] + O(1/t), m + 1 ≤ j ≤ mt.

(8)

To rigorously prove the existence of the steady-state degree distribution of the D-w.m.emodel, we need a limit theorem of difference equation.

Lemma 3 (limit theorem). For the following difference equation

at+1 − at + btat

ct= dt, (9)

if limt→∞ dt = l, ct+1 − ct = 1 and limt→∞ bt = b ≥ 0,

limt→∞

at

t= lim

t→∞(at+1 − at) =l

1 + b. (10)

Proof. The following known result is needed in the proof.

Stolz-Cesaro Theorem [13]. Let {yn} be a monotone increasing sequence with yn → ∞, wehave lim

n→∞xn

yn= l if lim

n→∞xn+1−xn

yn+1−yn= l, where −∞ ≤ l ≤ +∞.

We can now proceed with the proof. Let λn =1

1+bn, xn =

an

cnand yn = λn(an+1−an)+(1−λn) an

cn.

By the conditions of the Theorem, we have

yn − xn = λncn+1(xn+1 − xn), limn→∞ yn = lim

n→∞dn

1 + bn= λl, (11)

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D. H. Shi et al. / Physics Procedia 00 (2009) 1–9 6

where λ = limn→∞ λn = limn→∞ 11+bn= 1

1+b . Noting that cn+1 − cn = 1 and 1 + b(n) > 0 forsufficiently large n, without loss of generality, we assume that cn > 1 and λ(n) > 0. It is clearfrom the formula (11) that both yn and xn+1 are either larger or smaller than xn simultaneously,and furthermore, xn+1 is between yn and xn if λncn+1 > 1.

Obviously, for a given ε > 0, there is a positive integer N such that when n ≥ N we haveλcn+1 > 1, and from the formula (11)

λl − ε < yn < λl + ε.

Next, we prove that there is a positive integer N0 such that when N0 > N we have λl − ε <xN0 < λl + ε. We consider two cases.

Case 1: xN > λl + ε. In this case, either xN ≥ xN+1 ≥ · · · ≥ λl + ε or there is a N0 > N suchthat λl−ε < yN0−1 < xN0 < λl+ε < xN0−1. However, the former is not possible. On the one hand,when xN ≥ xN+1 ≥ · · · ≥ λl + ε holds, the limit limn→∞ xn = x exists and x > λl. On the otherhand, because an+1 − an =

yn

λn+

(1−λn)λn

an

cn, we have limn→∞(an+1 − an) = l + (1−λ)

λx < λl. Based

on this result and using Stolz-Cesaro Theorem, it implies that x = limn→∞ xn = limn→∞ an

cn=

limn→∞ an+1−an

cn+1−cn= limn→∞(an+1 − an) < λl. The two conclusions are contradictory. This shows

that there is a positive integer N0 such that λl − ε < xN0 < λl + ε.Case 2: xN < λl − ε. In this case, either xN ≤ xN+1 ≤ · · · ≤ λl − ε or there is a N0 > N such

that xN0−1 < λl−ε < xN0 < yN0−1 < λl+ε. However, the former is not possible. On the one hand,when xN ≤ xN+1 ≤ · · · ≤ λl − ε holds, the limit limn→∞ xn = x exists and x < λl. On the otherhand, because an+1 − an =

yn

λn+

(1−λn)λn

an

cn, we have limn→∞(an+1 − an) = l + (1−λ)

λx > λl. Based

on this result and using Stolz-Cesaro Theorem, we should have x = limn→∞ xn = limn→∞ an

cn=

limn→∞ an+1−an

cn+1−cn= limn→∞(an+1 − an) > λl. The two conclusions are contradictory. This shows

that there is a positive integer N0 such that λl − ε < xN0 < λl + ε.Because λl − ε < xN0 < λl + ε and xN0+1 is in the interval of yN0 and xN0 , we should have

λl − ε < xN0+1 < λl + ε. Thus, for all n ≥ N0, we have

λl − ε < xn < λl + ε.

This shows that limn→∞ xn = λl. Finally, taking the limit on both sides of yn = λn(an+1 − an) +(1 − λn) an

cn, and using limn→∞ yn = limn→∞ xn = λl, we get limt→∞(at+1 − at) = l

1+b . Again byStolz-Cesaro Theorem, we have limt→∞ at

t = limt→∞(at+1 − at).The proof of Theorem is completed.

Now we can prove the main result of this paper.Theorem 1. For the D-w.m.e model, the limit of the transient degree distribution P(k, t) exists

and is given by

P(k) =2m(m + 1)

k(k + 1)(k + 2), k = 1, 2, · · · . (12)

Proof. In the formula (8), for fixed j, let E[Nj(t)] = at,j2 = bt, t = ct and ( j−1)

2t E[Nj−1(t)] +δ jm + O(1/t) = dt. When j = m, since E[Nm−1(t)] = 0, all conditions in Lemma 3 are satisfied.Hence, limt→∞ P(m, t) = limt→∞

E[Nm(t)](t+1) = limt→∞

E[Nm(t)]t = 1

1+(m/2) =2

m+2 exists. Obviously, ifthe limit of P( j, t) exists, it also may imply that the limit of P( j + 1, t) exists. By mathematicalinduction, the limit of the transient degree distribution P(k, t) exists and the steady-state degreedistribution {P(k)} satisfy the following recursive relation.

P(k) =

{ 2m+2 , k = m(k−1)(k+2) P(k − 1), k = m + 1, · · · . (13)

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D. H. Shi et al. / Physics Procedia 00 (2009) 1–9 7

Finally, through recursive computations of P(k) until k = m + 1, i.e.,

P(k) = (k−1)(k+2) P(k − 1) = (k−1)

(k+2)(k−2)(k+1) P(k − 2)

=(k−1)(k+2)

(k−2)(k+1)

(k−3)k

(k−4)(k−1) · · · (m+2)

(m+5)(m+1)(m+4)

m(m+3) P(m),

we obtain formula (12). The proof of Theorem 1 is completed.

4. Application and conclusions

For comparison, we again apply the new approach to the LCD-m model.For the LCD-m model, when t = 1, the initial network consists of one vertex and m loop

edges. Then at every time step, we add a new vertex with m edges connecting to the existing net-work. At time t, the total degrees of the network are still 2mt, but the possible maximal degree of avertex is mt+m. Hence, the vector Markov chain of the network is N(t) = {Nm(t), · · · ,Nk(t), · · ·},where Nk(t) ≡ 0 for all k > mt + m. In addition, for large enough t, at time step t + 1, theprobability of the l-th edge connecting k−degree vertex is pkl =

k2mt+2(l−1)+1 =

k2mt+2(m−1)+1 +

o(1/t) = pkm + o(1/t), but the probability of the l-th edge connecting the new vertex itself isp0l =

12mt+2(l−1)+1 = O(1/t), where 1 ≤ l ≤ m. Thus, when m ≥ 2, the probability that a vertex

received more than an edge is also O(1/t). In the following arguments, we may also ignore thecases when a vertex connecting multiple edges and loops provided that there is the term O(1/t)in the result.

Let pkm =k

2mt+2(m−1)+1 and αkm = pkmNk(t) < 1, the condition distributions and expectationsof Ym(t) and Yk(t) for the LCD-m model are the same as the D-w.m.e model except αkm insteadof αk. Hence we have

{E[Ym(t)] = 1 − mpmmE[Nm(t)] + (1 − δm1)O(1/t),E[Yk(t)] = m

(p(k−1)mE[Nk−1(t)] − pkmE[Nk(t)]

)+ O(1/t).

(14)

Using the limit result of difference equation to formula (14) recursively, we obtain the sameconclusion as given in [4]. Obviously, our approach for proving stability of the LCD-m model issimpler than the method in [4].

We may point out that our approach does not need to assume that the vertex number Nk(t)with degree k is continuous and to derive the differential equation of Nk(t) by heuristic arguments.More importantly, it does not need to assume the existence of P(k). The approach not only hasclear physical meaning but also is simpler mathematically. This framework provides a rigoroustheoretical basis for the rate equation approach which has been widely applied to many problemsin complex networks, e.g., epidemic spreading and dynamic synchronization. We believe thevertex-number-evolving Markov chain will have important applications in the science network.

Appendix: The proof of the last equality in (7)Let αk−1 = a and αk = b, using binomial expansion, the last equality in (7) can be rewritten

as a combinatorial formula:m∑m−1

k=1 (−1)kCkm−1(akb − abk)

=

m∑

l=1

lml∑

j=1

m−l−2 j∑

k=0

k∑

s=0

(−1)kCl+ jm C j

m−l− jCkm−l−2 jC

sk(al+ j+sb j+k−s − a j+sbl+ j+k−s). (15)

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D. H. Shi et al. / Physics Procedia 00 (2009) 1–9 8

Proof. In order to prove formula (15), we need to compare coefficients of terms apbq in bothsides of (15). Obviously, there are only terms +apb, p > 1 and −abq, q > 1 in the left-handsaid. Because l + j + k − s ≥ 2, there are terms +apb, p > 1 only in the first summation of theright-hand said. By p = l + j + s, j + k − s = 1, 0 ≤ s ≤ k and 1 ≤ j, it implies that j = 1,s = k = p − l − 1 and l ≤ p − 1. Thus, the coefficients of the terms +apb, p > 1 in the firstsummation of the right-hand said are

fp1 =

p−1∑

l=1

l(−1)p−l−1Cl+1m C1

m−l−1Cp−l−1m−l−2Cp−l−1

p−l−1 = (−1)pmCpm−1

p∑

l=2

(−1)l (l − 1)p!l!(p − l)!

= (−1)pmCpm−1.

Hence, fp1 are the same as the coefficients of the terms +apb, p > 1 in the left-hand said.The reason why the last equality in the above expression holds is as follows. Using combi-

national formulas (−1)kCkn−1 =

∑kj=0(−1) jC j

n and∑n

j=1(−1) j+1 jC jn = 0, we obtain

∑pl=2(−1)l (l−1)p!

l!(p−l)! =∑p

l=2(−1)llClp −∑p

l=2(−1)lClp

=(−∑p

l=1(−1)l+1lClp +C1

p

)−(∑p−1

l=0 (−1)lClp + (−1)pCp

p +C1p − 1)

= 1 − (−1)p−1 − (−1)p = 1.

Because l + j + s ≥ 2, there are terms −abq, q > 1 only in the second summation of the right-hand said. By the same arguments, we obtain that the coefficients of both sides for the terms−abq, q > 1 are also equal. The remainder is to prove that the sum of other terms on the right-hand said equals zero.

Let k = p + q − l − 2, s = p − l − j, then the terms including apbq on the right-hand side are

(−1)p+qm∑

l=1

ml∑

j=1

(−1)llCl+ jm C j

m−l− jCp+q−l−2 jm−l−2 j [Cq− j

p+q−l−2 j −Cp− jp+q−l−2 j]a

pbq. (16)

Because p and q are symmetrical in (16), it is sufficient to consider cases p > q > 1. In the casep > q > 1. The restrictions for the first part of the right-hand side are

p + q − l − 2 j ≥ p − l − j ≥ 0, p + q − l − 2 j ≤ m − l − 2 j,

which means (l, j) must lie in the area D1 = {(l, j)|1 ≤ j ≤ q, 1 ≤ l ≤ p − j}. The restrictions forthe second part of the right-hand side are

p + q − l − 2 j ≥ p − j ≥ 0, p + q − l − 2 j ≤ m − l − 2 j,

which means (l, j) must lie in the area D2 = {(l, j)|1 ≤ j < q, 1 ≤ l ≤ q − j}.First we calculate the first part in (16),

(−1)p+q ∑(l, j)∈D1

(−1)llCl+ jm C j

m−l− jCp+q−l−2 jm−l−2 j Cq− j

p+q−l−2 japbq

=(−1)p+qm!

(m−p−q)!p!q!

∑(l, j)∈D1

(−1)llCl+ jp C j

qapbq

=(−1)p+qm!

(m−p−q)!p!q! [q∑

k=2

k−1∑j=1

(−1)k− j(k − j)CkpC j

q +p∑

k=q+1

q∑j=1

(−1)k− j(k − j)CkpC j

q]apbq.

Similarly we calculate the second part in (16),

(−1)p+q ∑(l, j)∈D2

(−1)llCl+ jm C j

m−l− jCp+q−l−2 jm−l−2 j Cp− j

p+q−l−2 japbq

=(−1)p+qm!

(m−p−q)!p!q!

q∑k=2

k−1∑j=1

(−1)k− j(k − j)CkqC j

papbq.

1764 D. Shi et al. / Physics Procedia 3 (2010) 1757–1765

D. H. Shi et al. / Physics Procedia 00 (2009) 1–9 9

A tiresome calculation we can derive the following results:

q∑k=2

k−1∑j=1

(−1)k− j(k − j)CkpC j

q +p∑

k=q+1

q∑j=1

(−1)k− j(k − j)CkpC j

q

= −p − pq∑

k=2

k∑j=2

(−1) j+kC j−1p−1Ck

q + qq∑

k=2

k∑j=2

(−1) j+kC jpCk−1

q−1;

q∑k=2

k−1∑j=1

(−1)k− j(k − j)CkqC j

p

= −q + qq∑

k=2

k−1∑j=0

(−1) j+kCk−1q−1C j

p − pq∑

k=2

k−1∑j=1

(−1) j+kCkqC j−1

p−1.

Thus, the summation of the corresponding terms of apbq for p > q > 1 on the right-hand side is

(−1)p+qm!(m−p−q)!p!q! (−1)p+qapbq{−p − p

q∑k=2

k∑j=2

(−1) j+kC j−1p−1Ck

q + qq∑

k=2

k∑j=2

(−1) j+kC jpCk−1

q−1+

+ q − qq∑

k=2

k−1∑j=0

(−1) j+kCk−1q−1C j

p + pq∑

k=2

k−1∑j=1

(−1) j+kCkqC j−1

p−1}.

However,

{−p − pq∑

k=2

k∑j=2

(−1) j+kC j−1p−1Ck

q + qq∑

k=2

k∑j=2

(−1) j+kC jpCk−1

q−1+

+q − qq∑

k=2

k−1∑j=0

(−1) j+kCk−1q−1C j

p + pq∑

k=2

k−1∑j=1

(−1) j+kCkqC j−1

p−1}

= −p −q∑

k=2

p!q!k!(k−1)!(p−k)!(q−k)! + p − pq + q

+ q[q−1∑k=0

(−1)kCkq−1 − 1] − pq[

q−1∑k=0

(−1)kCkq−1 − 1] +

q∑k=2

p!q!k!(k−1)!(p−k)!(q−k)!

= −p + p − pq + q − q + pq = 0,

which means the summation of the corresponding terms of apbq for p > q > 1 on the right-handside is zero. Hence, the equality (15) holds.

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